Syllabus based on AKTU for B.Tech students,

1. Engineering Physics I
Unit III (Diffraction)
Presentation By
Dr.A.K.Mishra
Professor
Jahangirabad Institute of Technology,
Barabanki
Email: akmishra.phy@gmail.com
Arun.Kumar@jit.edu.in
6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
1

2. Diffraction
• Huygens principle state that waves spread out after they
pass through slit.
• Bending of light at the corner of an obstacle.
• There are two type of diffraction
Fresnel diffraction
Distance of of screen from
aperture is finite.
Fraunhofer diffraction
Distance of screen from
the aperture is infinite.
6/26/2017
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JIT Jahangirabad
2

3. RESULTANT OF N HARMONICS
• Consider n simple harmonic vibrations having equal
amplitudes, periods and phases increasing in arithmetic
progression by an amount. To find the resultant amplitude of
such vibrations, a R polygon as shown in Figure 5.5 is
constructed. The closing side OP will give the resultant
amplitude R and a resultant phase. Resolving the amplitudes
parallel and perpendicular to first vibration. We get
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Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
3

4. Continue
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JIT Jahangirabad
4
2
1)-(n
Cosx
2
n
Sin2a
2
SinCos2R
)
2
1
-Sin(n
2
Sina
2
SinCos2R
)
2
3
-Sin(n-1)-Sin(n().......
2
3
Sin-
2
5
Sin()
2
Sin-
2
3
Sin(
2
sin2
2
SinCos2R
becomesequationaboveThe
B)-(Asin-B)(ASinBSin2cosAthatknow
1)-(ncos
2
2sin.......2cos
2
2sincos
2
2sin
2
sin2a
2
sincos2R
ly.respectivehavecanone
2
2cosby(2)eqnand
2
2sin(1)byeqngMultiplyin
.....(2)..........1)-sin(na.........sina0SinR
)1.....(..........1)-acos(n.......cosaaCos





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


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


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
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





















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





a
we
R

5. • Similarly
6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
5
becomes(5)eqnthen2nngagain takiAnasuch that
largeveryisnandsmallveryareandawhen
2
)1(
)6....(....................
2
1)-(n
tantan
getwe(3)by(4)eq
)5.........(....................
2
2
a
R
getwe(4)and(3Sin2)equationaddingand
)4(....................
2
Sin
2
1)-(n
Sinx
2
Sina
SinR
getwe
2
2Sinby(2)eqgmultiplyin
)3(....................
2
Sin
2
1)-(n
Cosx
2
Sina
Cos
n
2
22
2
n






















n
dividing
Sin
n
Sin
Squaring
n
Similarly
n
R

6. • We get
6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
6
Phaseresultantandamplituderesultantforexpressiontheis
22
1)-(n
ASin
R
small.very
Sin
naSin
2
2
2
which
n
and
R
n
Sin
aSin
R













7. Fraunhofer diffraction due to single slits
• S is the source and AB is the slit. Un deviated rays are focused
at O while diffracted rays through θ are focused at P. Path
beyond AN is same
Therefore path difference BN = AB sin θ = e sin θ
6/26/2017
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JIT Jahangirabad
7

8. • Phase difference =
Let n secondary wavelets between A & B therefore phase
difference between any two consecutive parts
6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
8
)sin(
2



e
22
)
Sin
(AIIntensity
na)(A
SinSin
na
sin
aR
n,largefor
2
sin
aR
sine
sine
Sin
sine
Sin
a
2
2
n
sin
aR
(Say))Sin(
21















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













A
n
nn
Sin
Sin
let
n
Sin
e
n

9. Condition for maxima and minima
• R is maximum when all the negative terms vanishes
6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
9
....)....................0,1,2,3,4.(mm
i.e0#but0Sin:minima
Amaximaprincipalcalledis0formaximumisp0
0
sine
0,
vanishestermsnegativetheallenmaximun whis
)....................
7!5!3!
-(1
)....................
7!
7
5!3!
-(
sin
A
2
642
53



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












For
R
AR
A
R

10. • Hence
6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
10
0atexceptmaximaprincipalthegives0
sin
tani.e
0sin-cosor0
sin
0)
sin-cos
)(
sin
2()
sin
(
0
d
dI
maximasecondaryFor
etc.minima..........fourth....third,second,first,ofdirectionthegives
.........)..........1,2,3.....(msin
sine
2
222


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




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
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AA
d
d
me
m

11. 6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
11
2
11
,
2
9
,
2
7
,
2
5
,
2
3
0,
tan
bygivenismaximasecondaryofposition





The

12. • Similarly the intensity of secondary maxima
6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
12
...................................
81
4
:
49
4
:
25
4
:
9
4
:1
it.ofsideeitheronintensitydecreasingofmaxima
secondaryAndintensityzeroofminimabyfollowed
maximacentralbrightofconsistpatternndiffractioThus
25
4
62
I
62
A
)
2
7
2
7
Sin
(A
9
4
22
I
22
A
)
2
5
2
5
Sin
(A
2222
2
2
0
2
22
2
2
2
0
2
22
1





AI
AI



13. Diffraction Grating
• A large number of parallel slits of equal width separated by
opaque spaces. Constructed equidistant line on a plane glass
plate with help of a diamond point .generally there are 10,000
to 15,000 lines per inch in a plane transmission grating .
6/26/2017
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JIT Jahangirabad
13

14. Diffraction Grating
• Let e be the width of each slit and d be the separation
between two consecutive slits then (e+d) is known as grating
element.
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JIT Jahangirabad
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














Sin
NSinSin
A'
Sin
NSinR
)
2
2
(
2
2N
SinR
R'
bygivenisPatamplituderesultantHence
.2differencephasecommon
havingwavesNofPatamplitudeResultant
(say)2Sind)(e
2
differencePhase
Sind)(eMSiss&sbetweendifferencePath
Sin
AReach wave,ofAmplitude
221





R
Sin

15. Diffraction Grating
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JIT Jahangirabad
15
0NSinifBut
slits.Ntodueintensityofondistributithe
)
Sin
NSin
(slit whilesingleto
dueintensitythegives)
Sin
(A
)
Sin
NSin
()
Sin
(A)2(R'I
bygivenisP
atintensityResultant
2
2
22











gives

16. Diffraction Grating
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Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
16























m0Sind)(e
m
Sind)(e
N
mN0SinBut
0NSinwhenminiumisIntensity
maxima.ncipal.......pri..........3rd,2nd,1stfor
........3.........2,,1nfornSind)(e
0,0nforn
sine
n
0SinmaximaPrincipalforN)
Sin
(AI
N
Sin
N
lim
Sin
NSin
lim
.ateindeterminis
Sin
NSin
Hence
22









 
N
d
d
Sin
d
d
nn

17. Diffraction Grating
6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
17
• Formation of spectra with a grating:
order.increasingwith
decreaseitorderzeroinedconcentratisIntensity
white.alwaysismax.principal
orderzero.henceofvalueallfor0=0,=nfor
n.oforder
differentfordifferentis,&d)+(eofvaluefixedafor
v.>randv>rwavelength
differentfordifferentis(n&d)+(eofvaluefixedFor
nsin)d(
th




 e

18. Diffraction Grating
• Condition for absent spectra:
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JIT Jahangirabad
18
absent.arespectra.order……2,4,6,then
d=eslittoequalisgratingtheofw idththeifHence
................2,4,6,8...
.....)1,2,3.....m(w here2mN
eqabovefromthenedIf1.
spectra.gratingisabsent
beorder tonthforconditiontheisThis
m
m
n
e
de
(2)&(1)eqnFrom
.(2)......)...1,2,3.....(mmSine
slitsingletodueminimaofDirection
.......(1)..........nSin)(









e
de
n
de



19. Diffraction Grating
2. If d=2e then , N =3m (where m = 1,2,3,……………..)
= 3,6,9……….. Spectra are missing.
Maximum number of order possible in grating spectra :
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JIT Jahangirabad
19
obtained.beorder willfirst
onlyincidencenormalforhence2
2
N
then,2d)(eifthatshows
)(90Sin)(
Sin)(
nSin)(
max
max
















This
dede
N
de
N
de

20. Diffraction Grating
• Examples: The tracks of a compact disc act as a diffraction
grating, producing a separation of the colors of white light.
• The illustration shows the
hydrogen spectrum.The hydrogen
gas in a thin glass tube is excited by
an electrical discharge and the
spectrum can be viewed through
the grating.
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JIT Jahangirabad
20

21. Dispersive power of a Grating
• Defined as the rate of change of the angle of diffraction with the wave length of light.
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JIT Jahangirabad
21
spectrum.normalasknownisspectrumaSuch
.ddhence1,Cossmall,isIf
region.in violetan thatgreater thisregionredthe
in thedispersionthe,therefore,spectrumgivenainvrAs
high.is
d
powerdispersive
andsmallerbewillCosthenlarge,isofvalueIf
CostoalproportionInversely(3)
d)(eelementgratingtoalproportioninversely(2)
norderthetoalproportionDirectely(1)
grating.theofpowerdispersivetheis
)1.........(..........
Cosd)(e
nd
n
d
Cosd)(e
getwe,t.r.it w.DifratingnSind)(e
isequationgratingtheknowweas,
d
asexpressedis

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


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d
This
d
d
d
it

22. Resolving power of an optical instruments
• The resolving power of an instrument is its ability to reveal
the minor details of the object under examination.
• When an object is placed in front of a convex lens at a point
beyond its focus, a real and inverted image of the object is
formed as shown in the figure below:
• The ratio of the size of the image
to the size of the object is called
magnification.
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JIT Jahangirabad
22

23. Resolving Power
• The resolving power of an instrument is the ability of an
optical instrument to distinguish (or resolve) two closely
spaced objects.
• Resolving power of the Telescope : The resolving power of
telescope is defined as the minimum angle between two
objects of equal magnitude (intensity) to see them just as
separate is called the resolving power of the telescope and
is proportional to the separation of the wavelength of light
being observed to the diameter of the telescope. Thus, the
larger the diameter, the smaller the minimum angular
separation and the higher the resolving power.
6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
23

24. Resolving Power for a Microscope
• The resolving power of a microscope is defined as the smallest separation
at which two separate objects can be distinguished. The resolving power
of a microscope is a function of the wavelength of light used and the
numerical aperture (NA) which is characteristic of the lens system and the
medium between the lens and the specimen.
• Rayleigh Criterion of Resolution:
In the 19th century, Lord Rayleigh invented a criteria for determining when
two light sources were
distinguishable from each other, or resolved.
According to the criteria, two point sources are
considered just resolved (just distinguishable
enough from each other to recognize two sources)
if the center of the diffraction pattern of one is
directly overlapped by the first minimum of the
diffraction pattern of the other
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JIT Jahangirabad
24

25. Resolving Power of Grating
• The resolving power of diffraction grating is is ability to form
separate spectral lines for wavelengths very close together.
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Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
25
element.gratingtheisd)(e
nSind)(ehavethen we
directionin theformedismaximaprincipalnlet
grating.onnormallyincidentlightthebedandLet
.gthat wavelenresolvedjustbecanthatdifference
avelengthsmallest wtheisd,
d
bygivenisIt
th

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
where
where


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




26. Resolving Power of Grating
• -
6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
26
 dand 

d
d



ofmin&
dofmax.
first
nth

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n
maxima.principal
,3rd2nd,1stthegivesvaues
thesebecause.nN,2N,3N....N,0
valuethehavingintegeranismand
gratingonrulingofNo.theis
)1.....(mSinD)(eN
isminimaforeqn
.ddirectionIn the
formedismaximanthe
oadjacent tminfirstLet•
th
N
grating
then

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

27. Resolving Power of Grating
• -
6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
27
grating.ndiffractioofpowerResolving
.......(4)....................nN
dnN
dnNnNnN
)dn(
N
1nN
get)we(3and(2)eqCompairing
......(3)).........dn()d(Sind)(e
.thenddirectionin theobtainedalsois
dofmaximanthwhengratingby theresolvejustare
d&hwavelengtthecriterionRayleighto
)2.......()
N
1Nn
()d(Sind)(eor
1)N(n)d(Sind)(eN
1)N(nmforobtained
bewill)increasing(ifdirectionin themaximaprincipalntheoadjacent tminimumfirstThe th
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
d
d
According

28. Resolving Power of Grating
• -
6/26/2017
Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
28
powervexdispersiapertureTotalgratingofpowerResolving
)6.........(..........
d
d
.cosd)(eNNn
d
asexpressed
bemaygratingtheofpowerresolvingthe
)5......(..........
cosd)(e
n
d
d
bygivenisgrtingtheofpowerdispersiveAs
spectrum.theoforderthe
andgratingon therulingtheofno.totaltheofproduct
thetoequalisgratingtheofpowerresolvinthe



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






Therfore
Thus